# Prove that the center of a group is a normal subgroup

Let $G$ be a group. We define $H$ where $H$ is the center of/centralizer of $G$:

$$H=\{h\in G| \forall g\in G: hg=gh\}.$$

Prove that $H$ is a (normal) subgroup of $G$.

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No, it was my exam question. Just finnished it. I think I managed it up to part to prove that it is normal. But I couldn't prove that it is a subgroup. – Rok Kralj Sep 20 '11 at 12:15
Just curious: what does it mean to prove that it is normal without proving that it is a subgroup? – ShreevatsaR Sep 20 '11 at 12:28
@ShreevatsaR Let $S$ be a subset of a group $G$. We say that $S$ is a normal subset of $G$ if for all $s\in S$ and $g\in G$, we have $g^{-1}sg\in S$. In particular, a subset $S$ of $G$ is a normal subgroup of $G$ if and only if $S$ is a subgroup of $G$ and $S$ is a normal subset of $G$. – Amitesh Datta Sep 20 '11 at 13:15
@AmiteshDatta: Thanks. – ShreevatsaR Sep 20 '11 at 13:25

As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b \in H$, we know that $\forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab \in H$.

Now we only have to show that every $h \in H$ has an inverse and we are done. Let $h \in H$, we know that $\forall g \in G: gh = hg$, thus \begin{align*}h^{-1}(gh)h^{-1} &= h^{-1}(hg)h^{-1}\\ h^{-1}g(hh^{-1}) &= (h^{-1}h)gh^{-1}\\ h^{-1}(ge) &= (eg)h^{-1}\\ h^{-1}g &= gh^{-1} \end{align*}

Which implies that $h^{-1} \in H$.

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@ Srivatsan, thanks for the english lesson :) – sxd Sep 20 '11 at 12:37

I have two different solutions. These are probably too fancy for this problem, but they might be interesting.

For the first solution, define the map $f: G \rightarrow \operatorname{Aut}(G)$ by $g \mapsto \phi_g$, where $\phi_g(x) = gxg^{-1}$ for all $x \in G$. The map $f$ is a homomorphism and $\operatorname{Ker}(f) = Z(G)$. Thus $Z(G)$ is a normal subgroup since the kernel of a homomorphism is always a normal subgroup. The image $\operatorname{Im}(\phi)$ is called the inner automorphism group of $G$ and is denoted $\operatorname{Inn}(G)$.

For the second solution, recall that for a subgroup $H \leq G$ the normal core of $H$ in $G$ is defined as

$$\operatorname{core}_G(H) = \bigcap_{g \in G} gHg^{-1}$$

The subgroup $\operatorname{core}_G(H)$ is always a normal subgroup. This can be seen directly or by noticing that it is the kernel of the coset action induced by $H$. For any conjugacy class $C$ of $G$, let $t_c \in C$. Let $\mathscr{C}$ be the family of all conjugacy classes of $G$. Then \begin{align*} Z(G) &= \bigcap_{g \in G} C_G(g) \\ &= \bigcap_{C \in \mathscr{C}} \bigcap_{t \in C} C_G(t_c) \\ &= \bigcap_{C \in \mathscr{C}} \bigcap_{g \in G} C_G(gt_cg^{-1}) \\ &= \bigcap_{C \in \mathscr{C}} \bigcap_{g \in G} gC_G(t_c)g^{-1} \\ &= \bigcap_{C \in \mathscr{C}} \operatorname{core}_G(C_G(t_c)) \\ \end{align*}

Since $Z(G)$ is the intersection of normal subgroups, it is a normal subgroup.

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i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.

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